Exercicios Calculo (189)

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205 
 
𝑓(−3) = 3𝑎(−3) − 7𝑏 = −9𝑎 − 7𝑏. 
lim
𝑥→−3−
𝑓(𝑥) = lim
𝑥→−3−
(3𝑥 + 6𝑎) = −9+ 6𝑎. 
lim
𝑥→−3+
𝑓(𝑥) = lim
𝑥→−3+
(3𝑎𝑥 − 7𝑏) = −9𝑎 − 7. 
 
𝐿𝑜𝑔𝑜, 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = −3: 
 
−9+ 6𝑎 = −9𝑎 − 7𝑏 
15𝑎 + 7𝑏 = 9 
 
2. 𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑑𝑎𝑑𝑒 𝑑𝑒 𝑓 𝑒𝑚 𝑥 = 3: 
lim
𝑥→3
𝑓(𝑥) = 𝑓(3) ⟹ {
lim
𝑥→3−
𝑓(𝑥) = 𝑓(3)
𝑒
lim
𝑥→3+
𝑓(𝑥) = 𝑓(3)
 
 
𝑓(3) = 3𝑎(3) − 7𝑏 = 9𝑎 − 7𝑏. 
lim
𝑥→3−
𝑓(𝑥) = lim
𝑥→−3−
(3𝑎𝑥 − 7𝑏) = 9𝑎 − 7𝑏. 
lim
𝑥→3+
𝑓(𝑥) = lim
𝑥→3+
(𝑥 − 12𝑏) = 3− 12𝑏. 
 
𝐿𝑜𝑔𝑜, 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 3: 
 
3 − 12𝑏 = 9𝑎− 7𝑏 
9𝑎 + 5𝑏 = 3 
 
𝐶𝑜𝑚 𝑖𝑠𝑠𝑜, 𝑡𝑒𝑚𝑜𝑠: {
15𝑎 + 7𝑏 = 9
9𝑎 + 5𝑏 = 3
↷ {
45𝑎 + 21𝑏 = 27
−45𝑎 − 25𝑏 = −15
↷ {
45𝑎 + 21𝑏 = 27
−4𝑏 = 12
. 
𝑆𝑜𝑙𝑢çã𝑜: 𝑏 = −3 𝑒 𝑎 = 2. 
 
𝑃𝑎𝑟𝑎 𝑒𝑠𝑡𝑒𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑎 𝑒 𝑏 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 ℝ. 
 
𝐶𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎 𝑑𝑒 𝑟𝑎𝑖𝑜 1 𝑐𝑒𝑛𝑡𝑟𝑎𝑑𝑎 𝑛𝑎 𝑜𝑟𝑖𝑔𝑒𝑚: 𝑥2 + 𝑦2 = 1 
𝑅𝑒𝑡𝑎 𝑦 = 𝑎𝑥 + 𝑏: 𝑦 = 2𝑥 − 3. 
 
𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑠𝑒 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑖𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑒𝑛𝑡𝑟𝑒 𝑎 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎 𝑒 𝑎 𝑟𝑒𝑡𝑎: 
 
𝑥2 + (2𝑥 − 3)2 = 1 
𝑥2 + 4𝑥2 −12𝑥 + 9 = 1 
5𝑥2 − 12𝑥 + 8 = 0 
∆= 144 − 160 = −16